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Commit dfe3d56

Browse files
feat: update 7 leetcode
1 parent b57efc0 commit dfe3d56

8 files changed

+290
-0
lines changed

‎leetcode/169.majority-element.ts‎

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/*
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* @lc app=leetcode id=169 lang=typescript
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*
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* [169] Majority Element
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*/
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// @lc code=start
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/**
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* @description: 哈希表的方式
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* 时间复杂度 O(n)
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* 空间复杂度 O(n)
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* @param {number[]} nums
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* @return {number}
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*/
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function majorityElement(nums: number[]): number {
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let hash = {};
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for (const num of nums) {
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hash[num] = hash[num] || 0;
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hash[num]++;
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if (hash[num] > nums.length / 2) return num;
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}
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}
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/**
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* @description: BM 算法的思路
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* 时间复杂度 O(n)
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* 空间复杂度 O(1)
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* @param {number[]} nums
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* @return {number}
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*/
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function majorityElement2(nums: number[]): number {
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let count = 0;
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let candi;
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for (const num of nums) {
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// 前面的字符抵消了
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if (count === 0) {
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candi = num;
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}
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// 如果相同,就继续增加,如果不同则减少
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count += candi === num ? 1 : -1;
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}
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return candi;
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}
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// @lc code=end
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/*
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* @lc app=leetcode id=416 lang=typescript
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*
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* [416] Partition Equal Subset Sum
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*/
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// @lc code=start
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/**
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* @description: 0-1 背包的扩展,这里用回溯实现;但实际更推荐使用动态规划
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* @param {number[]} nums
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* @return {boolean}
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*/
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function canPartition(nums: number[]): boolean {
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const sum = nums.reduce((pre, cur) => pre + cur, 0);
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const half = sum / 2;
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// 说明总和是奇数,取一半有小数点,也就不存在子数组和相等的可能
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if (half !== Math.floor(half)) return false;
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let hasPartied = false;
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let memory = new Array(nums.length).fill(null).map(() => ({}));
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/**
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* @description: 回溯解法
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* @param {number[]} arr
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* @param {number} index
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* @param {number} currentSum
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* @return {void}
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*/
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function backtrack(arr: number[], index: number, currentSum: number): void {
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if (index === arr.length) {
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// 判断如果能得到总和为一半的数组的情况,那么就是存在相等和的子数组
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if (currentSum === half) hasPartied = true;
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return;
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}
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// !使用缓存降低时间复杂度,提高效率
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if (memory[index][currentSum]) return;
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memory[index][currentSum] = true;
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backtrack(arr, index + 1, currentSum);
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if (currentSum + nums[index] <= half) {
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backtrack(arr, index + 1, currentSum + nums[index]);
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}
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}
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backtrack(nums, 0, 0);
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return hasPartied;
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}
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// @lc code=end

‎leetcode/494.target-sum.ts‎

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/*
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* @lc app=leetcode id=494 lang=typescript
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*
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* [494] Target Sum
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*/
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// @lc code=start
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/**
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* @description: 回溯的解法,也可以使用动态规划
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* @param {number} nums
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* @param {number} target
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* @return {number}
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*/
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function findTargetSumWays(nums: number[], target: number): number {
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let count = 0;
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/**
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* @description: 定义回溯函数
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* @param {number[]} arr
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* @param {number} target
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* @param {number} index
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* @param {number} currentSum
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* @return {void}
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*/
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function backtrack(arr: number[], target: number, index: number, currentSum: number): void {
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if (index === arr.length) {
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currentSum === target && count++;
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return;
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}
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backtrack(arr, target, index + 1, currentSum + arr[index]);
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backtrack(arr, target, index + 1, currentSum - arr[index]);
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}
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backtrack(nums, target, 0, 0);
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return count;
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}
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// @lc code=end

‎leetcode/509.fibonacci-number.ts‎

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/*
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* @lc app=leetcode id=509 lang=typescript
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*
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* [509] Fibonacci Number
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*/
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// @lc code=start
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/**
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* @description: 动态规划
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* @param {number} n
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* @return {number}
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*/
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function fib(n: number): number {
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const res = [];
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res[0] = 0;
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res[1] = 1;
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for (let i = 2; i <= n; i++) {
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// !核心的递推公式
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res[i] = res[i - 1] + res[i - 2];
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}
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return res[n];
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}
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// @lc code=end
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/*
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* @lc app=leetcode id=674 lang=typescript
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*
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* [674] Longest Continuous Increasing Subsequence
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*/
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// @lc code=start
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/**
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* @description: 贪心算法
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* 时间复杂度 O(n)
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* 空间复杂度 O(1)
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* @param {number} nums
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* @return {numer}
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*/
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function findLengthOfLCIS(nums: number[]): number {
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let max = 1;
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let current = 1;
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for (let i = 1; i < nums.length; i++) {
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if (nums[i - 1] < nums[i]) {
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current += 1;
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max = Math.max(max, current);
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} else {
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current = 1;
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}
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}
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return max;
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}
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// @lc code=end

‎leetcode/70.climbing-stairs.js‎

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return memory[n] = n === 1 || n === 2 ? n : fibonacci(n - 1) + fibonacci(n - 2);
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}
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})();
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/**
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* 进阶版,翻译
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* @param {*} num
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* @returns
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*/
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function num2str(num) {
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const numStr = num + ''
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let first = 1
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let second = 0
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if ([1, 2].includes(numStr[0]) && numStr[i] < 6) {
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second = 2
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} else {
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second = 1
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}
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for (let i = 1; i < numStr.length; i++) {
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if ([1, 2].includes(+numStr[i - 1]) && +numStr[i] < 6) {
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let middle = second
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second = first + second
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first = middle
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}
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}
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return second
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}
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/*
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* @lc app=leetcode id=728 lang=typescript
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*
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* [728] Self Dividing Numbers
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*/
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// @lc code=start
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/**
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* @description: 判断每一位是否可以被除
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* @param {number} left
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* @param {number} right
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* @return {number[]}
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*/
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function selfDividingNumbers(left: number, right: number): number[] {
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let start = left;
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const res = [];
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while (start <= right) {
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if (start < 10) {
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res.push(start);
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start++;
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continue;
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}
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let isDivided = true;
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for (const str of String(start)) {
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if (start % Number(str) !== 0) {
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isDivided = false;
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continue;
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}
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}
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isDivided && res.push(start);
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start++;
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}
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return res;
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}
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// @lc code=end

‎leetcode/77.combinations.ts‎

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/*
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* @lc app=leetcode id=77 lang=typescript
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*
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* [77] Combinations
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*/
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// @lc code=start
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/**
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* @description: 回溯算法的实践
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* @param {number} n
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* @param {number} k
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* @return {number[][]}
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*/
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function combine(n: number, k: number): number[][] {
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const res = []; // 保存结果的数组
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const currentList = []; // 每次组合的数组
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/**
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* @description: 递归函数用于计算保存
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* @param {number} n
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* @param {number} k
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* @param {number} startIndex
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* @return {void}
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*/
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function combineArr(n: number, k: number, startIndex: number): void {
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if (currentList.length === k) {
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// !注意避免引用类型的坑
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res.push(currentList.slice());
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return;
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}
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// !剪枝的方式提高效率,在判断当前数组内元素已经到达设定上限时不再继续执行
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for (let i = startIndex; i <= n - (k - currentList.length) + 1; i++) {
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// for (let i = startIndex; i <= n; i++) {
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currentList.push(i);
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combineArr(n, k, i + 1); // 下一个数字进行判断
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currentList.pop(); // 空出来用于进入下一轮判断
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}
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}
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combineArr(n, k, 1);
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return res;
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}
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// @lc code=end

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